document.write( "Question 562910: A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B; if the accident has occurred 50 miles further on, it would have been 2 hours late. Find the distance from A to B and the original velocity of the train? \n" ); document.write( "
Algebra.Com's Answer #364939 by ankor@dixie-net.com(22740)\"\" \"About 
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A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B;
\n" ); document.write( " if the accident has occurred 50 miles further on, it would have been 2 hours late.
\n" ); document.write( "Find the distance from A to B and the original velocity of the train?
\n" ); document.write( ":
\n" ); document.write( "Let s = normal speed of the train
\n" ); document.write( "then
\n" ); document.write( ".6s = speed after accident
\n" ); document.write( ":
\n" ); document.write( "d = distance from A to B
\n" ); document.write( "then
\n" ); document.write( "\"d%2Fs\" = normal travel time of the train from A to B
\n" ); document.write( "\"d%2Fs\" + 3 = travel time after 50 mi accident
\n" ); document.write( "\"d%2Fs\" + 2 = travel time after 100 mi accident
\n" ); document.write( ":
\n" ); document.write( "50 mi accident travel time equation
\n" ); document.write( "\"50%2Fs\" + \"%28%28d-50%29%29%2F%28.6s%29\" = \"d%2Fs\" + 3
\n" ); document.write( "multiply by .6s to clear the denominators, results
\n" ); document.write( ".6(50) + (d-50) = .6d + .6s(3)
\n" ); document.write( "30 + d - 50 = .6d + 1.8s
\n" ); document.write( "d - 20 = .6d + 1.8s
\n" ); document.write( "d - .6d - 1.8s = 20
\n" ); document.write( ".4d - 1.8s = 20
\n" ); document.write( ":
\n" ); document.write( "100 mi accident travel time equation
\n" ); document.write( "\"100%2Fs\" + \"%28%28d-100%29%29%2F%28.6s%29\" = \"d%2Fs\" + 2
\n" ); document.write( "multiply by .6s, results
\n" ); document.write( ".6(100) + (d-100) = .6d + .6s(2)
\n" ); document.write( "60 + d - 100 = .6d + 1.2s
\n" ); document.write( "d - 40 = .6d + 1.2s
\n" ); document.write( "d - .6d - 1.2s = 40
\n" ); document.write( ".4d - 1.2s = 40
\n" ); document.write( ":
\n" ); document.write( "Subtract the 1st equation from the above equaton
\n" ); document.write( ".4d - 1.2s = 40
\n" ); document.write( ".4d - 1.8s = 20
\n" ); document.write( "----------------
\n" ); document.write( "+.6s = 20
\n" ); document.write( "s = \"20%2F.6\"
\n" ); document.write( "s = 33\"1%2F3\" mph is the normal train speed
\n" ); document.write( "Find d
\n" ); document.write( ".4d - 1.2(33.33) = 40
\n" ); document.write( ".4d - 40 = 40
\n" ); document.write( ".4d = 40 + 40
\n" ); document.write( ".4d = 80
\n" ); document.write( "d = 80/.4
\n" ); document.write( "d = 200 mi is the distance
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "See if the that checks out in the 100 mi accident equation
\n" ); document.write( "\"100%2F33.33\" + \"%28%28200-100%29%29%2F%28.6%2833.33%29%29\" = \"200%2F33.33\" + 2
\n" ); document.write( "3 + \"100%2F20\" = 6 + 2
\n" ); document.write( "3 + 5 = 6 + 2
\n" ); document.write( ":
\n" ); document.write( "We can say: normal speed = 33\"1%2F3\" mph; distance is 200 mi
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